Author Topic: The Beauty of Math (Read 4481 times)

Serge · « **Reply #20 on:** October 16, 2008, 04:31:24 PM »

Quote from: Flying_Chao;17511

How did you get " a - 1 = a + 1 " ?

I divided both sides by (a - 1).

No, sh** dammit, I made a mistake. I'll fix the image in a sec.

Serge · « **Reply #21 on:** October 16, 2008, 04:33:39 PM »

Here, fixed.

Flying_Chao · « **Reply #22 on:** October 16, 2008, 04:36:07 PM »

That's better.

Naryar · « **Reply #23 on:** October 16, 2008, 04:49:20 PM »

I don't see how the hell do you jump from a-1=(a-1)*(a+1) to a-1=a+1...

And you forgot something important: the equation a^2 =1 admits TWO values of a: a=1and a =-1. So, a=1 => a^2=1 goes one way (the opposite isn't true) but all of the others goes both way (mathematical symbol <=>).

With a=-1 the thing works.

I bet that can be solved by imaginary numbers but I'm too lazy to use them now.

Yeah, I'm a math nerd.

Serge · « **Reply #24 on:** October 16, 2008, 04:55:12 PM »

Quote from: Naryar;17521

I don't see how the hell do you jump from a-1=(a-1)*(a+1) to a-1=a+1...

And you forgot something important: the equation a^2 =1 admits TWO values of a: a=1and a =-1. So, a=1 => a^2=1 goes one way (the opposite isn't true) but all of the others goes both way (mathematical symbol <=>).

With a=-1 the thing works.

I bet that can be solved by imaginary numbers but I'm too lazy to use them now.

Yeah, I'm a math nerd.

The first error was fixed, my bad, I am tired, refresh the image.

As for the confusion with a^2... I am nowhere telling that sqrt(a^2) = 1, so I do not need absolute values or deal with multiple cases.

Gigafrost · « **Reply #25 on:** October 16, 2008, 07:51:09 PM »

An easy question:

Solve this problem without using the number 2 or anything that 2, 2.5, 1.5, or 7 is divisible by. Yet the numbers that create the product must be a whole number, equal to 2 and greater than 1.

1+1 = ?

Sage · « **Reply #26 on:** October 16, 2008, 07:55:50 PM »

square root of 4.

you said '2 is divisible by' instead of 'divisble by 2'

otherwise just 5-3

Gigafrost · « **Reply #27 on:** October 16, 2008, 07:57:22 PM »

Nu uh (smiley face)... Fixed =)

andrewm0304 · « **Reply #28 on:** October 16, 2008, 08:37:26 PM »

Haha, thats a good one Serge

man manu · « **Reply #29 on:** October 17, 2008, 01:09:06 PM »

-1+3?

Sage · « **Reply #30 on:** October 17, 2008, 02:04:56 PM »

@man manu: 2. not even math really.

heres a question:

theres an undefined line and a line with the slope zero. they are perpendicular to each other. whats the slope of the undefined line?

Serge · « **Reply #31 on:** October 17, 2008, 02:09:23 PM »

Quote from: Sage;17624

@man manu: 2. not even math really.

heres a question:

theres an undefined line and a line with the slope zero. they are perpendicular to each other. whats the slope of the undefined line?

The undefined line cannot be the graph of a function, since in a function no argument can result in more than one value (where in this case, there is an infinite numbers of values).

Sage · « **Reply #32 on:** October 17, 2008, 02:25:11 PM »

incorrect

Naryar · « **Reply #33 on:** October 17, 2008, 04:00:21 PM »

No, he's right.

This can't have any slope because if we consider it in a classic two-dimensional space (cartesian coordinates) with Ox and Oy as vectors, that (vertical) line would have an equation of x=k (k being a constant and y being free).

Problem, you can't have a slope (or can't get a derivative for being more precise) from such a function.

Common sense would say the value of the slope is fixed and infinite, but this is a math abomination.

So you can't have any slope on there, excepted if you're tricking us with 3D space.

Sage · « **Reply #34 on:** October 17, 2008, 04:26:41 PM »

well, from what i know, an undefined slope= a vertical line. if a slope = 0 then its horizontal, meaning they form a perpendicular. the answer was in the question.

the slope is undefined, because in the slope formula the answer would be something divided by zero, which is impossible, making it a vertical line.

R0B0SH4RK · « **Reply #35 on:** October 17, 2008, 08:57:25 PM »

Basically, since slope is rise of the line over the run, having a completely vertical line is mathematically impossible from my understanding. The slope would be infinity divided by zero. In a distance over time graph (one measuring speed), such a line would mean that the object is everywhere in the universe at the same time.

Quote from: Serge;17509

(Image removed from quote.)

There is a flaw in the fourth line here. a-1 = (a-1)(a+1) should equal a-1 = a^2-1
which is 0=0, not 1=2 :P

Gigafrost · « **Reply #36 on:** October 18, 2008, 12:22:18 AM »

What is the square root of infinity?

Sage · « **Reply #37 on:** October 18, 2008, 12:23:35 AM »

Quote from: Gigafrost;17691

What is the square root of infinity?

pi.

Naryar · « **Reply #38 on:** October 18, 2008, 02:44:45 AM »

Infinity too!

Serge · « **Reply #39 on:** October 18, 2008, 04:00:51 AM »

Quote from: R0B0SH4RK;17685

Basically, since slope is rise of the line over the run, having a completely vertical line is mathematically impossible from my understanding. The slope would be infinity divided by zero. In a distance over time graph (one measuring speed), such a line would mean that the object is everywhere in the universe at the same time.

There is a flaw in the fourth line here. a-1 = (a-1)(a+1) should equal a-1 = a^2-1
which is 0=0, not 1=2 :P

There is nothing wrong with this line. The error is somewhere else.